fix opt/8634

[Andrew Pinski]

But really the math is K*n*O(n) = O(n^2) for any K>0 aka you call the 
function K*n times, and the function takes O(n) time. So really it is 
still O(n^2).

Thanks,
Andrew Pinski


On Wednesday, Apr 9, 2003, at 00:31 US/Eastern, Gerald Pfeifer wrote:

> On Tue, 8 Apr 2003, Geoff Keating wrote:
>> If you take a linear operation (that is, O(n)), and then you perform
>> it K*n times for some fixed constant K, then the program is O(n*n),
>> which is quadratic.
>
> Sorry, but this is certainly not how big-O notation is supposed to 
> work:
> O(K·n) = O(n) for any fixed K > 0, by definition.
>
> Gerald
> -- 
> Gerald "Jerry"   pfeifer@dbai.tuwien.ac.at   
> http://www.pfeifer.com/gerald/
>
>